Does a filament lamp still have resistance when no current flows?
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7 Answers
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Let's be clear about what the resistance is. It's a measure of the opposition to current flow in an electrical circuit. For many materials, the current $I$ through the material is approximately proportional to the voltage $V$ applied across it: $$V\propto I;$$ $$V=IR,$$ where $R$ is the resistance of the wire (or material). Note that resistance depends on length, cross-section area, temperature, etc. Thus, it's better to define a quantity $\rho$ that depends on the material that satisfies $$R=\rho\frac{l}{A};$$ for more, see here.
That's the basis we need.
Does in a filament lamp still has resistance when no current flows?
The answer is Yes!, as we define the resistance as the ability of an object to resist a current. If there is no current in the wire, it doesn't mean that it loses this ability. To measure this ability, however, we generally need to flow current through the filament.
It may be Ohm's law (the second expression above) that's bugging you. Note that if $I=0$, then $V=0$—but this doesn't imply that $R =0$.
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$\begingroup$ I would say the first expression above, $V\propto I;$ is Ohm's and the second expression V=IR is a definition of resistance. $\endgroup$– M. EnnsCommented Oct 31, 2020 at 22:51
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$\begingroup$ @M.Enns V is not proportional to I if R varies in accordance with I (e.g. because the filament is getting hot and releasing light). $\endgroup$ Commented Nov 1, 2020 at 8:51
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$\begingroup$ @nick012000 It still has a resistance though - it's just that the resistance changes with temperature. At no point is there no resistance. $\endgroup$– GrahamCommented Nov 1, 2020 at 14:10
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$\begingroup$ @M.Enns That's a distinction without a difference. Ohm's law states that
VandIare directly proportional, meaning there is a constant factor of proportionality. We've called thatR. So both expressions say exactly the same thing and are both therefore descriptions of Ohm's law. $\endgroup$ Commented Nov 1, 2020 at 22:31 -
$\begingroup$ "If there is no current in the wire, it doesn't mean that it loses this ability." Ha! we don't know that! Whenever I don't look the toys in my room are having a party, after all! Then I switch the light on -- everything quiet. It's amazing how good they are at that. $\endgroup$ Commented Nov 2, 2020 at 13:18
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Yes, since resistance is an objective physical property of a material/object, whose existence is independent on whether we are observing it or not. In the same way as an object has mass, even when it is floating in an open space.
This is a rather serious philosophical question, which usually goes under the headline: "Does a falling tree make a sound, even if there is no one to hear it?"
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The filament of a light bulb has a resistance which varies with current (and temperature). If you measure the current as a function of voltage, you can plot the resistance as a function of current. On this plot, in the limit as the current approaches zero, the curve has an intercept that gives the resistance at zero current.
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$\begingroup$ It varies with current directly, not just temperature? So you're saying the voltage vs. current curve isn't linear even for momentary changes in current, before the filament has significant time to heat or cool. Or if we hold temperature constant by cooling the surrounding fluid while raising current, so we can actually measure it without inductance / reactance effects. I'm not sure that's right. If that's not what you meant, probably different phrasing that attributes resistance to temperature directly, which has a time-lagged dependence on average current^2. $\endgroup$ Commented Nov 1, 2020 at 8:29
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$\begingroup$ But yes, agreed with the last sentence, and the experiment you describe. $\endgroup$ Commented Nov 1, 2020 at 8:29
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1$\begingroup$ There is very little time lag between increasing the current and seeing an increase in temperature. $\endgroup$ Commented Nov 1, 2020 at 21:01
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1$\begingroup$ Yes, it's fast, but not instantaneous; not fast on time time-scale of the peaks and valleys of a 60Hz sine wave, and certainly not if you used higher frequency AC power. Given the option when there's no lack of clarity, it's best to phrase things in ways that are actually true and imply the correct physics. You might want to say "resistance which varies with temperature (which varies with average power)". $\endgroup$ Commented Nov 1, 2020 at 21:35
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Not only does it have resistance, it also has current, and non-zero voltage...even when it's unplugged.
That is because of the statistical nature of the dissipative process of "resistance" turning current into heat and the fluctuation-dissipation theorem, which applies to any dissipative system. In this case, it means that thermal fluctuations in the electrons leads to a rapidly varying current that is always present (and is called Johnson noise, or Johnson-Nyquist noise, or Nyquist noise, or thermal noise...but definitely not shot noise, which is different).
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There are perfectly good answers here, but I don't see any of them clearly stating what you'll observe if you try to measure the resistance of an incandescent lightbulb with most multi-meters.
At low currents (and a hand-held multi-meters typically measure resistance by feeding a very low current/voltage), most incandescent bulbs are basically short-circuits. It may actually say $0\Omega$. (see comments) In fact, even if your meter can't detect it, there is some small resistance, much more than the truly tiny resistance of copper household-wiring. This means that when current does flow, nearly all of that power shows up as heat in the filament. Once the filament gets hot enough, the tungsten looses a lot of it's conductance; the resistance goes up.
When the filament of a 100W bulb is glowing hot enough to give off the amount of light we expect of it, its resistance is ~$100\Omega$ (because $P=\frac{V^2}{R}$), but at room temperature it's a lot lower.
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2$\begingroup$ From a 120V power supply, power = V^2 / R. For a 100W bulb, solving for R = 120V^2 / 100W, R = 144 ohms. Perhaps you have a bulb that can draw 100W from a 12V supply, at 8.3 amps? That sounds like a disaster for inrush current when switched on, but yes it would have 1.44 ohm resistance at full temperature, and fractions of an ohm at room temp. $\endgroup$ Commented Nov 1, 2020 at 8:35
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$\begingroup$ I found a 50W and a 150W incandescent bulb (120V) in a drawer: at room temp, I measured ~21 ohms for the 50W, and 6.6ohms for the 150. Minus probably 0.5 ohms at the contacts; estimated by putting both probes onto one part of the bulb, the previous numbers were the raw meter readings, not trying to correct for measurement error; a lightbulb socket -> bulb connection would also have some resistance, but hopefully less than I got digging in sharp probes past some corrosion. $\endgroup$ Commented Nov 1, 2020 at 8:44
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$\begingroup$ @PeterCordes: Fixed, and thanks! I had a ghost whispering the wrong answer in my ear last night. $\endgroup$ Commented Nov 1, 2020 at 13:58
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1$\begingroup$ Your ~0Ω measurement claim for the room-temp resistance still seems unlikely, and you forgot to fix that. As I said, even a 150W bulb (for 120V) measured 6.6Ω on my multimeter. You definitely do not want an inrush current of over 120 amps when you first flip the light switch. Most digital multi-meters can measure down to at least tenths of an ohm in their lowest range (e.g. 0 - 200Ω), so reading 0 is totally implausible there (even if you did subtract the minimum measurement when you short the contacts to each other). My relatively cheap hobbyist ones can, including my dad's from the 80s. $\endgroup$ Commented Nov 1, 2020 at 21:59
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$\begingroup$ @PeterCordes I don't doubt that most incandescent bulbs measure significantly more than zero on a typical MM; however, you seem to be arguing from the premise that (positive voltage) / (zero resistance) = (infinite current), i.e., that bulbs are perfect resistors, which is not true. When you connect a voltage source to a superconductor the current does not instantaneously become infinite. $\endgroup$– trentCommented Nov 2, 2020 at 0:12
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In a sense, yes and no.
Ohm's law defines resistance as a ratio between the voltage and the current. It is considered a constant for a particular conductor over a wide range of voltages, currents, conductors, conditions AND accuracy requirements. It may as well not be constant.
Postulating U=0 and I=0 makes the resistance an undefined value. Then again, for most real-world cases, resistance has well-defined "limit" (in semi-math sense) near U->0 and I->0. That's why we don't have any issue the resistance the absense of voltage and current.
For most practical purposes, filament lamps cannot be assumed having constant resistance. Their resistance vary as much as 10 times depending on the applied voltage and for how long it is applied. This is an important consideration when powering filament lamps.
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$\begingroup$ Indeed, not only is it important, it is downright useful. There are plenty of examples of taking advantage of this property, using a filament lamp as a current limiter. If the current gets too high, the resistance of the lamp goes up, limiting the current. It's easy to show the maximum current that can flow, regardless of what the rest of the circuit does, using just this lamp, and you can use that information to do things like sizing your wires to make sure they don't melt. $\endgroup$ Commented Oct 31, 2020 at 23:48
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$\begingroup$ Ohm's Law does not define resistance as the ratio between current and voltage. Ohm's Law says that in certain circumstances the voltage is proportional to the current. $\endgroup$– M. EnnsCommented Nov 1, 2020 at 15:57
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I would like to give a theoretical and different approach for Ohmic Resistors
Ohm's law is given as $$V=IR$$ Note that we have three variables. However for a normal resistor the value of R is fixed (it does vary with temperature as current flows though it but let's neglect that).
This means that no matter what value of $I$ you place in the equation, R will still be the same. When you think about $I=0$ is not a very different case. Even though current becomes zero (or tends to zero for very low Voltage) , the resistance is still the same.
With all this said, the question in itself is rather vague. Resistance is a property to oppose electric current through an object. However if there is no current through the object, then it is not opposing it. So does that mean that it has lost its ability to oppose for this case? We know that it has not lost this property forever but it is also true that the resistance is not actively resisting anything.
So this does not seem like a "yes/no' question and probably the answer depends on a person's perception.
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$\begingroup$ Normally we use capital $I$ for current when we're using capital $V$. Lower-case $i$ does get used for current when it's a function of time, not just an unknown. It seems inconsistent to use that here when we're presumably talking about some unknown DC or rms value. V/I uppercase/lowercase convention? $\endgroup$ Commented Nov 1, 2020 at 22:04
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$\begingroup$ @PeterCordes Thank you for revision. $\endgroup$ Commented Nov 2, 2020 at 1:14